AMC 10 · 2014 · #21
Grade 8 geometry-2dTrapezoid ABCD has parallel sides AB of length 33 and CD of length 21. The other two sides are of lengths 10 and 14. The angles A and B are acute. What is the length of the shorter diagonal of ABCD?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Trapezoid $ABCD$ has parallel sides $AB = 33$ and $CD = 21$, and its two slanted legs measure $10$ and $14$. Angles $A$ and $B$ (at the long base) are acute. Find the length of the shorter of the two diagonals $AC$ and $BD$.
Givens: $AB \parallel CD$ with $AB = 33$ and $CD = 21$; The two legs have lengths $10$ and $14$; Angles $A$ and $B$ at the long base are acute; Answer choices: (A) $10\sqrt{6}$, (B) $25$, (C) $8\sqrt{10}$, (D) $18\sqrt{2}$, (E) $26$
Unknowns: The length of the shorter diagonal of $ABCD$
Understand
Restated: Trapezoid $ABCD$ has parallel sides $AB = 33$ and $CD = 21$, and its two slanted legs measure $10$ and $14$. Angles $A$ and $B$ (at the long base) are acute. Find the length of the shorter of the two diagonals $AC$ and $BD$.
Givens: $AB \parallel CD$ with $AB = 33$ and $CD = 21$; The two legs have lengths $10$ and $14$; Angles $A$ and $B$ at the long base are acute; Answer choices: (A) $10\sqrt{6}$, (B) $25$, (C) $8\sqrt{10}$, (D) $18\sqrt{2}$, (E) $26$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #13 Convert to Algebra
A trapezoid with two given bases and two given legs is begging for coordinates, so Tool #1 (Draw a Diagram) is primary: lay the long base $AB$ on the $x$-axis and drop perpendiculars from the top corners. Tool #4 (Introduce a Variable) names the two horizontal offsets $p$ and $q$ and the height $h$. Tool #7 (Identify Subproblems) turns each leg into its own right triangle via the Pythagorean theorem. Tool #13 (Convert to Algebra) then solves the little system for $p$, $q$, $h$, after which the two diagonals are just distances between corner points.
Execute — Answer: B
6.G.A.3 Step 1 Put the trapezoid on a grid
- Lay the long base on the $x$-axis: $A = (0, 0)$ and $B = (33, 0)$.
- The short base $CD = 21$ sits at some height $h$, directly above the middle stretch.
- Let $D = (p,\, h)$ be above the left part and $C = (p + 21,\, h)$ be above it $21$ units to the right, so $DC$ stays parallel to $AB$ with length $21$.
💡 Once both bases are horizontal, every corner has clean coordinates and slanted sides become right triangles you can measure.
8.EE.C.7 Step 2 Name the two base offsets
- The bottom base is $33$ and the top base is $21$, so the leftover $33 - 21 = 12$ is split between the two overhangs.
- Call the left overhang $p$ (the horizontal run of leg $AD$) and the right overhang $q$ (the horizontal run of leg $BC$).
- Since the acute base angles keep both feet inside $AB$, both are positive and they must fill the gap: $p + q = 12$.
💡 The extra length of the long base has to go somewhere — it splits into the two slanted overhangs at the ends.
8.G.B.7 Step 3 Pythagoras on each leg
- Each leg is the hypotenuse of a right triangle with horizontal run ($p$ or $q$) and vertical rise $h$.
- Take the leg of length $10$ on the left and the leg of length $14$ on the right (the other assignment is just a mirror image and gives the same diagonals).
- Then $p^2 + h^2 = 10^2 = 100$ and $q^2 + h^2 = 14^2 = 196$.
💡 A slanted side over a flat floor always makes a right triangle, so its length ties the horizontal run and the height together.
8.EE.C.8 Step 4 Solve for the offsets and height
- Subtract the two Pythagorean equations to kill $h^2$: $q^2 - p^2 = 196 - 100 = 96$.
- Factor the left side as $(q - p)(q + p)$ and use $q + p = 12$, so $(q - p)(12) = 96$, giving $q - p = 8$.
- Combine with $q + p = 12$ to get $q = 10$ and $p = 2$.
- Back-substitute: $h^2 = 100 - p^2 = 100 - 4 = 96$.
💡 Subtracting the two equations erases the shared height, leaving a simple pair of linear facts about $p$ and $q$.
8.G.B.8 Step 5 Measure both diagonals
- Now every corner is known: $A=(0,0)$, $B=(33,0)$, $D=(2,\,h)$, $C=(23,\,h)$ with $h^2 = 96$.
- Diagonal $AC$ has horizontal run $23$ and vertical run $h$, so $AC^2 = 23^2 + h^2 = 529 + 96 = 625$, giving $AC = 25$.
- Diagonal $BD$ has horizontal run $33 - 2 = 31$, so $BD^2 = 31^2 + 96 = 961 + 96 = 1057$, giving $BD \approx 32.5$.
- The shorter diagonal is $AC = 25$, so the answer is $\textbf{(B)}$.
💡 A diagonal is just the straight-line distance between two known corners — one more right triangle from run and rise.
6.G.A.3 Lay the long base on the $x$-axis: $A = (0, 0)$ and $B = (33, 0)$. The short bas 8.EE.C.7 The bottom base is $33$ and the top base is $21$, so the leftover $33 - 21 = 12$ 8.G.B.7 Each leg is the hypotenuse of a right triangle with horizontal run ($p$ or $q$) 8.EE.C.8 Subtract the two Pythagorean equations to kill $h^2$: $q^2 - p^2 = 196 - 100 = 9 8.G.B.8 Now every corner is known: $A=(0,0)$, $B=(33,0)$, $D=(2,\,h)$, $C=(23,\,h)$ with Review
Reasonableness: The height works out to $h = \sqrt{96} = 4\sqrt6 \approx 9.8$, comfortably less than both legs $10$ and $14$, which is required since a leg is always at least as long as the height it spans — good. The diagonal $AC = 25$ reaches from a bottom corner across to the far top corner; $25$ is longer than the longer leg $14$ and shorter than the bottom base $33$, exactly the range you expect for a diagonal. $BD \approx 32.5$ is clearly the longer one because $D$ sits almost directly above $A$, leaving $BD$ to span nearly the whole base. So the shorter diagonal is $25$, matching (B). The distractors $10\sqrt6 \approx 24.5$, $8\sqrt{10} \approx 25.3$, $18\sqrt2 \approx 25.5$, and $26$ all cluster near $25$ to punish arithmetic slips, but only the exact $625 = 25^2$ lands on a clean value.
Alternative: Instead of subtracting equations, split the trapezoid with the perpendicular from $C$. The diagonal $AC$ closes a right triangle whose horizontal leg is $A$-to-foot-of-$C = 23$ and whose vertical leg is the height. Finding the height once (drop both perpendiculars, use the $10$-$12$-$14$ overhang split $p=2$, $q=10$ to get $h^2 = 96$) and then applying the Pythagorean theorem to that single triangle gives $AC = \sqrt{23^2 + 96} = 25$ directly, and a quick check shows $BD$ is longer, so no full coordinate bookkeeping is strictly needed.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing $A=(0,0)$, $B=(33,0)$, and the top corners $D=(p,h)$, $C=(p+21,h)$ so the trapezoid becomes coordinate data.)8.EE.C.7Solve linear equations in one variable (Writing the base-difference relation $p + q = 33 - 21 = 12$ between the two horizontal overhangs.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Turning each leg into a right triangle: $p^2 + h^2 = 100$ and $q^2 + h^2 = 196$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the system $q + p = 12$, $q - p = 8$ (from subtracting the Pythagorean equations) to get $p = 2$, $q = 10$, $h^2 = 96$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing the diagonals $AC^2 = 23^2 + 96 = 625$ and $BD^2 = 31^2 + 96 = 1057$ from the corner coordinates.)
⭐ Stand the trapezoid on the $x$-axis, split each slanted leg into a right triangle, and the extra $33 - 21 = 12$ of base splits as $2$ and $10$; then diagonal $AC = \sqrt{23^2 + 96} = 25$ is the short one.
⭐ Stand the trapezoid on the $x$-axis, split each slanted leg into a right triangle, and the extra $33 - 21 = 12$ of base splits as $2$ and $10$; then diagonal $AC = \sqrt{23^2 + 96} = 25$ is the short one.
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